A coffee with temperature $LaTeX: \displaystyle 170^\circ$ is left in a room with temperature $LaTeX: \displaystyle 72^\circ$ . After 5 minutes the temperature of the coffee is $LaTeX: \displaystyle 164^\circ$ , what is the temperature of the coffee after 14 minutes?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 72+(170-72)e^{kt}= 72+98e^{kt}$ . Using the point $LaTeX: \displaystyle (5, 164)$ gives $LaTeX: \displaystyle 164= 72+98e^{k(5)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{46}{49}=e^{5k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{46}{49} \right)}}{5}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 72+98e^{\frac{\ln{\left(\frac{46}{49} \right)}}{5}t}$ and simplifying gives $LaTeX: \displaystyle T = 98 \left(\frac{46}{49}\right)^{\frac{t}{5}} + 72$ . Using $LaTeX: \displaystyle t = 14$ gives $LaTeX: \displaystyle T =98 \left(\frac{46}{49}\right)^{\frac{14}{5}} + 72\approx 154^\circ$